Problems with Coffee | Abdelouahab Mostafa

Showing $\int_0^1\tan^{-1}\frac{f(x)}{1+f(x)}\frac{dx}x=\frac\pi8\ln\frac{\pi^2}8$ , where $f(x)=\frac1\pi(\tanh^{-1}x-\tan^{-1}x)$ , with real analysis

Show that: I found this question [Here][1] on MSE. [Here][2] is the solution obtained by means of complex analysis. I found this question on MathOverflow [Here][3], but no answers

Problem 210 min

Show that : $\int^1_0 \frac{f(x^2)}{\sqrt{1-x^2}}dx =\frac{\pi^2}{16}$

nice proplem with cofffe in this night

Problem 310 min

evaluate : $\int_0^{\infty}\frac{\arctan\left({\frac{\sqrt{4x^2 +2x\sqrt{4x^2+1}}}{\sqrt{4x^2+1}-x}}\right)}{1+x^2} dx=\frac{\pi^2}{6}$

evaluate :

Problem 410 min

prove that : $\int^{\infty}_0 \frac{\sin\left({\frac{2}{\pi}x^2}\right)}{\sinh^2(2x)} dx =\frac{1}{8}$

A clean definite integral problem involving a quadratic sine phase and the hyperbolic sine function.

Show that :∫01f(x2)1−x2dx=π216\int^1_0 \frac{f(x^2)}{\sqrt{1-x^2}}dx =\frac{\pi^2}{16}∫01​1−x2​f(x2)​dx=16π2​

evaluate : ∫0∞arctan⁡(4x2+2x4x2+14x2+1−x)1+x2dx=π26\int_0^{\infty}\frac{\arctan\left({\frac{\sqrt{4x^2 +2x\sqrt{4x^2+1}}}{\sqrt{4x^2+1}-x}}\right)}{1+x^2} dx=\frac{\pi^2}{6} ∫0∞​1+x2arctan(4x2+1​−x4x2+2x4x2+1​​​)​dx=6π2​

prove that : ∫0∞sin⁡(2πx2)sinh⁡2(2x)dx=18\int^{\infty}_0 \frac{\sin\left({\frac{2}{\pi}x^2}\right)}{\sinh^2(2x)} dx =\frac{1}{8}∫0∞​sinh2(2x)sin(π2​x2)​dx=81​

Show that : $\int^1_0 \frac{f(x^2)}{\sqrt{1-x^2}}dx =\frac{\pi^2}{16}$

evaluate : $\int_0^{\infty}\frac{\arctan\left({\frac{\sqrt{4x^2 +2x\sqrt{4x^2+1}}}{\sqrt{4x^2+1}-x}}\right)}{1+x^2} dx=\frac{\pi^2}{6}$

prove that : $\int^{\infty}_0 \frac{\sin\left({\frac{2}{\pi}x^2}\right)}{\sinh^2(2x)} dx =\frac{1}{8}$